Answer :
Sure! Let's solve the problem step-by-step:
We're tasked with calculating the change in length, [tex]\(\Delta L\)[/tex], using the formula:
[tex]\[
\Delta L = L_0 \times \alpha \times \Delta T
\][/tex]
Where:
- [tex]\(L_0\)[/tex] is the initial length, given as 90 (unspecified units, but the calculation is unitless).
- [tex]\(\alpha\)[/tex] is the coefficient of linear expansion, given as [tex]\(11 \times 10^{-6}\)[/tex].
- [tex]\(\Delta T\)[/tex] is the change in temperature, given as 39.3 degrees.
Now, let's perform the calculation:
1. Start by writing down the formula:
[tex]\[
\Delta L = 90 \times (11 \times 10^{-6}) \times 39.3
\][/tex]
2. Multiply the initial length ([tex]\(L_0\)[/tex]), the coefficient of linear expansion ([tex]\(\alpha\)[/tex]), and the temperature change ([tex]\(\Delta T\)[/tex]):
[tex]\[
\Delta L = 90 \times 0.000011 \times 39.3
\][/tex]
3. Calculate the result:
- First, multiply [tex]\(90 \times 0.000011 = 0.00099\)[/tex].
- Then, multiply [tex]\(0.00099 \times 39.3 = 0.038907\)[/tex].
Thus, the change in length, [tex]\(\Delta L\)[/tex], is approximately 0.038907.
We're tasked with calculating the change in length, [tex]\(\Delta L\)[/tex], using the formula:
[tex]\[
\Delta L = L_0 \times \alpha \times \Delta T
\][/tex]
Where:
- [tex]\(L_0\)[/tex] is the initial length, given as 90 (unspecified units, but the calculation is unitless).
- [tex]\(\alpha\)[/tex] is the coefficient of linear expansion, given as [tex]\(11 \times 10^{-6}\)[/tex].
- [tex]\(\Delta T\)[/tex] is the change in temperature, given as 39.3 degrees.
Now, let's perform the calculation:
1. Start by writing down the formula:
[tex]\[
\Delta L = 90 \times (11 \times 10^{-6}) \times 39.3
\][/tex]
2. Multiply the initial length ([tex]\(L_0\)[/tex]), the coefficient of linear expansion ([tex]\(\alpha\)[/tex]), and the temperature change ([tex]\(\Delta T\)[/tex]):
[tex]\[
\Delta L = 90 \times 0.000011 \times 39.3
\][/tex]
3. Calculate the result:
- First, multiply [tex]\(90 \times 0.000011 = 0.00099\)[/tex].
- Then, multiply [tex]\(0.00099 \times 39.3 = 0.038907\)[/tex].
Thus, the change in length, [tex]\(\Delta L\)[/tex], is approximately 0.038907.
